What are the solutions to the equation (x+1)^2=2x^2 ?

The solutions to the equation (x+1)^2=2x^2 are x=1-sqrt(2),x=1+sqrt(2)

Find the zeros of (x+1)^2=2x^2

The zeros of (x+1)^2=2x^2 are x=1-sqrt(2),x=1+sqrt(2)

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(x + 1)^{2} = 2 x^{2}

x = 1 - 2, x = 1 + 2

Decimal Notation

x = - 0.41421 \dots, x = 2.41421 \dots

Solution steps

(x + 1)^{2} = 2 x^{2}

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

x^{2} + 2 x + 1 = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 1 = 22

x_{1, 2} = \frac{- 2 \pm 2 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 2}{2 ( - 1 )}, x_{2} = \frac{- 2 - 2 2}{2 ( - 1 )}

x = \frac{- 2 + 2 2}{2 ( - 1 )} : 1 - 2

x = \frac{- 2 - 2 2}{2 ( - 1 )} : 1 + 2

The solutions to the quadratic equation are:

x = 1 - 2, x = 1 + 2

0 = - x^{2}

7 x^{2} + b x + c = 0

5 v^{2} + 30 v = 0

3 x^{2} + x - 7 = 0

m^{2} - 2 m - 5 = 0

What are the solutions to the equation (x+1)^2=2x^2 ?
The solutions to the equation (x+1)^2=2x^2 are x=1-sqrt(2),x=1+sqrt(2)
Find the zeros of (x+1)^2=2x^2
The zeros of (x+1)^2=2x^2 are x=1-sqrt(2),x=1+sqrt(2)